# Functions

• Apr 2nd 2011, 10:54 AM
ikislav
Functions
Hello :)

The problem states:

IF $f(\frac{1}{x})=x^2+\frac{1}{x^3}$ AND $g(x)= \frac{1}{2}(f(x)+f(-x))$ THEN $g(x)=?$

So, I got that $f(x)=\frac{1}{x^2}+\frac{x^3}{1}$ and that $f(-x)=\frac{1}{x^2}-\frac{x^3}{1}$

That gives me $g(x)=\frac{1}{2}*\frac{2}{x^2}=\frac{1}{x^2}$

Is this correct? I had huge problems with this as I was always getting $g(x)=\frac{1}{2}*0$... then I figured my mistake: $f(-x)$ is not equal to $(-f(x))$
• Apr 2nd 2011, 11:25 AM
veileen
"Is this correct?" Yes, it is.
• Apr 3rd 2011, 05:03 AM
HallsofIvy
If f is any function, we can define the "even part" of f(x) to be $f_e(x)= \frac{f(x)+ f(-x)}{2}$ and the "odd part" of f(x) to be [tex]f_o(x)= \frac{f(x)- f(-x)}{2}[/itex]. $f_e$ is an "even function", f(-x)= f(x), $f_o(x)$ is an "odd functin", f(-x)= -f(x), and their sum is $f_e(x)+ f_o(x)= f(x)$.

If f is an even function to start with, its "odd part" will be 0 and if f is an odd function, its "even part" will be 0. Rational functions that involve only even powers of x are "even functions" and rational functions that involve only odd powers of x are "odd functions.

Here, you were asked to find the even part of the function $f(x)= \frac{1}{x^2}+ x^3$ which is, as you finally found, $\frac{1}{x^2}$. Its odd part is, of course, $x^3$.